Environmental Engineering Reference
In-Depth Information
J
(
q
) has its maximum, to a nearby commensurable value
Q
c
. Hence
the hexagonal anisotropy couples the helical magnetic structure to the
lattice, and it may induce continuous or abrupt changes of the ordering
wave-vector as a function of temperature, as discussed, for instance, by
Bak (1982). In Ho, 12
Q
0
is close to 4
π/c
, and the hexagonal anisotropy
is large at low temperatures. Experimental investigations have shown
that a number of commensurable values of
Q
are stabilized in this sys-
tem, as we shall discuss in more detail in the last section of this chapter.
2.1.4 Longitudinally ordered phases
When
B
2
is negative, as in Er and Tm,
χ
ζζ
(
Q
) is the component of the
susceptibility which diverges at the highest temperature, and the high-
temperature expansion predicts that 2
A
ζ
−J
(
Q
) vanishes at a critical
temperature determined by
(
Q
)
1
)
B
2
/k
B
T
N
.
k
B
T
N
3
−
5
−
2
)(
J
+
2
J
(
J
+1)
J
(
J
(2
.
1
.
31)
Just below this temperature, only the component
σ
ζ
at the wave-vector
Q
is non-zero and, from the free energy expansion (2.1.24),
∂f/∂σ
ζ
=0
determines the relative magnetization as
σ
ζ
(
Q
)=
σ
Q
=
J
1
/
2
2
A
ζ
3
J
2
B
ζζ
−
(
Q
)
.
(2
.
1
.
32)
The free energy is independent of the phase
ϕ
=
ϕ
ζ
,soweset
ϕ
=0. If
we add another Fourier component with
q
=
±
Q
:
=
Jσ
Q
cos (
Q
·
R
i
)+
Jσ
q
cos (
q
·
R
i
+
ϕ
)
J
iζ
(2
.
1
.
33)
then, if
m
Q
±
n
q
is different from a reciprocal lattice vector, where
m
and
n
are integers and
m
+
n
=
±
4, the free energy is
J
2
{
J
4
B
ζζ
3
σ
4
Q
+3
σ
q
+12
σ
2
Q
σ
q
+4
σ
3
Q
σ
q
δ
q
,±
3
Q
cos
ϕ
+4
σ
Q
σ
q
δ
3
q
,±
Q
cos 3
ϕ
.
(2
.
1
.
34)
σ
q
+
8
f
=
4
σ
2
Q
+
2
A
ζ
−J
(
Q
)
}
{
2
A
ζ
−J
(
q
)
}
1
This result shows that, if
q
=3
Q
or
q
=
3
Q
, there is an extra fourth-
order contribution to the free energy (
q
→−
q
represents the same
structure with
ϕ
→−
ϕ
). Of these two special cases, the one where
q
=3
Q
is the most interesting, because the extra term is linear in
σ
3
Q
.
This means that the third harmonic appears simultaneously with the
basic Fourier component at
Q
.
Minimizing the free energy given by
(2.1.34), we find
J
2
B
ζζ
J
(
Q
)
−J
(3
Q
)
σ
3
Q
ϕ
=
ϕ
+
π,
σ
3
Q
=
(2
.
1
.
35
a
)
;
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